How come this code generates the output of 11?

Question

INT I=4;
I=++I+I++;
cout<<I;

Answer 1

Hi,

Please refer to http://en.cppreference.com/w/cpp/language/operator_precedence for the precedence and associativity of C++ operators.

Basically from your expression
let,
I = 4,
x = ++I (4+1 = 5) = 5,
y = I++ (5+1 = 6) = 6,

I = (x) + (y)
= 5 + 6
= 11

Answer 2

@Ken Kaneki and @ATR110,

Pardon me if i'm wrong but isn't the operator's precedence for post-increment and pre-increment is higher than addition and subtraction?

Answer 3

@Corsair Alex In c++, pre-increment takes precedence over addition. Post-increment comes last.

Answer 4

@Corsair Alex

Pre-increment have preference over operators, post-increment not.But in this case the result is the same if you ignore the preference.

Answer 5

l initially is 4, the first increment operator makes it 5, so the second line becomes 5 + 5, and there's another increment operator at the end, which adds 1 to 5 + 5. 11.

Answer 6

You can split the code by this way:

1) Start -> I = 4

2) Operation: (a + b)

a) ++I -> I = 5

b) I++ -> 5 (At this moment I equals 5, only catch the current value of I, then it will increment in one more (++).

3) a = 5
b = 5
I = 5+5 = 10
But you have a ++, so it will be 11 when you print it.

Answer 7

thanks guys especially for the ones who split the expression and explained

Answer 8

but then ... why when I did :
int b=1;
b=b++;
cout<<b;
I obtained 1 ? :/